An eight turn coil encloses an elliptical area having a major axis of 40.0 cm an

Question

An eight turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm (Fig. P19.23). The coil lies in the plane of the page and has a 6.00 A current flowing clockwise around it. If the coil is in a uniform magnetic field of 2.08 ? 10-4 T, directed toward the left of the page, what is the magnitude of the torque on the coil? (Hint: The area of an ellipse is A = ?ab, where a and b are, respectively, the semimajor and semiminor axes of the ellipse.)... I need help... I only have two tries left

Explanation / Answer

A=pi*a*b =pi*(0.4/2)*(0.3/2) =0.094 m^2

T=NiABsin(x)

T =8*6*0.094*2.08*10^-4*sin90

T=9.41*10^-4 N-m

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